Simplify the following expression and state the condition under which the simplification is valid. You can assume that $a \neq 0$. $r = \dfrac{15a - 20}{-5} \times \dfrac{-3}{21a^2 - 28a} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $r = \dfrac{ (15a - 20) \times -3 } { -5 \times (21a^2 - 28a) } $ $ r = \dfrac {-3 \times 5(3a - 4)} {-5 \times 7a(3a - 4)} $ $ r = \dfrac{-15(3a - 4)}{-35a(3a - 4)} $ We can cancel the $3a - 4$ so long as $3a - 4 \neq 0$ Therefore $a \neq \dfrac{4}{3}$ $r = \dfrac{-15 \cancel{(3a - 4})}{-35a \cancel{(3a - 4)}} = -\dfrac{15}{-35a} = \dfrac{3}{7a} $